Section 12.2

Direct Summation of Finite Sums and Infinite Series

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Problem 5003

If uν=vν+1−vνu_{\nu } = v_{\nu +1} - v_{\nu }, then we have the identity:

∑ν=0n−1uν=vn−v0. \sum _{\nu =0}^{n-1} u_{\nu } = v_{n} - v_{0}.

This remark and similar related techniques give the solution to the following problems. Prove the identity:

∑ν=1nν2=n(n+1)(2n+1)6. \sum _{\nu =1}^{n} \nu ^{2} = \frac{n(n+1)(2n+1)}{6}.
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Problem 5004

Prove the identity ∑ν=1nν3=(n(n+1)2)2\sum_{\nu =1}^{n} \nu^{3} = \left(\frac{n(n+1)}{2}\right)^{2}.

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Problem 5005

Prove the identity 12+32+52+…+(2n−1)2=4n3−n31^{2}+3^{2}+5^{2}+\ldots +(2n-1)^{2} = \frac{4n^{3}-n}{3}.

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Problem 5006

Prove the identity 13+33+53+…+(2n−1)3=2n4−n21^{3}+3^{3}+5^{3}+\ldots +(2n-1)^{3} = 2n^{4}-n^{2}.

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Problem 5007

Prove the identity ∑ν=1n−1ν2(n−ν)2=n5−n30\sum_{\nu =1}^{n-1} \nu^{2}(n-\nu )^{2} = \frac{n^{5}-n}{30}.

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Problem 5008

Prove the identity 14−24+34−…+(−1)n−1n4=(−1)n−1n4+2n3−n21^{4}-2^{4}+3^{4}-\ldots +(-1)^{n-1}n^{4} = (-1)^{n-1}\frac{n^{4}+2n^{3}-n}{2}.

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Problem 5009

Prove the identity s5+s7=2s32s_{5}+s_{7} = 2s_{3}^{2}; sν=1ν+2ν+…+nνs_{\nu } = 1^{\nu }+2^{\nu }+\ldots +n^{\nu }.

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Problem 5010

Using Euler's formulas, find the sum: 1+cos⁡x+cos⁡2x+…+cos⁡(n−1)x1+\cos x+\cos 2x+\ldots +\cos (n-1)x.

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Problem 5011

Using Euler's formulas, find the sum: cos⁡x+cos⁡3x+…+cos⁡(2n−1)x\cos x+\cos 3x+\ldots +\cos (2n-1)x.

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Problem 5012

Using Euler's formulas, find the sum: sin⁡x+sin⁡2x+…+sin⁡nx\sin x+\sin 2x+\ldots +\sin nx.

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Problem 5013

Using Euler's formulas, find the sum: sin⁡x−sin⁡2x+…−sin⁡2nx\sin x-\sin 2x+\ldots -\sin 2nx.

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Problem 5014

Using Euler's formulas, find the sum: sin⁡2x−sin⁡23x+…+(−1)n−1sin⁡2(2n−1)x\sin^{2}x-\sin^{2}3x+\ldots +(-1)^{n-1}\sin^{2}(2n-1)x.

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Problem 5015

Using Euler's formulas, find the sum: xsin⁡a+x2sin⁡2a+…+xn−1sin⁡(n−1)ax\sin a+x^{2}\sin 2a+\ldots +x^{n-1}\sin (n-1)a.

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Problem 5016

Prove the equality x+2x2+3x3+4x4+…=x(1−x)2x+2x^{2}+3x^{3}+4x^{4}+\ldots = \frac{x}{(1-x)^{2}}; ∣x∣<1\left|x\right| < 1.

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Problem 5017

Prove the equality 1−4x+9x2−16x3+…=1−x(1+x)31-4x+9x^{2}-16x^{3}+\ldots = \frac{1-x}{(1+x)^{3}}; ∣x∣<1\left|x\right| < 1.

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Problem 5018

Prove the identity 1a(a+1)+1(a+1)(a+2)+…+1(a+n−1)(a+n)=1a−1a+n\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\ldots +\frac{1}{(a+n-1)(a+n)} = \frac{1}{a}-\frac{1}{a+n}.

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Problem 5019

Prove the identity

1a(a+1)(a+2)+1(a+1)(a+2)(a+3)+…+1(a+n−1)(a+n)(a+n+1)= \frac{1}{a(a+1)(a+2)}+\frac{1}{(a+1)(a+2)(a+3)}+\ldots +\frac{1}{(a+n-1)(a+n)(a+n+1)} = =12a(a+1)−12(a+n)(a+n−1). = \frac{1}{2a(a+1)} - \frac{1}{2(a+n)(a+n-1)}.
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Problem 5020

Prove the identity

1a(a+p)+1(a+1)(a+p+1)+…+1(a+n−1)(a+p+n−1)= \frac{1}{a(a+p)}+\frac{1}{(a+1)(a+p+1)}+\ldots +\frac{1}{(a+n-1)(a+p+n-1)} = =1p(1a+1a+1+…+1a+p−1)−1p(1a+n+1a+n+1+…+1a+n+p−1). = \frac{1}{p}\left(\frac{1}{a}+\frac{1}{a+1}+\ldots +\frac{1}{a+p-1}\right) - \frac{1}{p}\left(\frac{1}{a+n}+\frac{1}{a+n+1}+\ldots +\frac{1}{a+n+p-1}\right).
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Problem 5021

Prove the equality 11⋅3+13⋅5+15⋅7+…=12\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\ldots = \frac{1}{2}.

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Problem 5022

Prove the equality 11⋅4+14⋅7+17⋅10+…=13\frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+\ldots = \frac{1}{3}.

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Problem 5023

Prove the equality 11⋅4+12⋅5+13⋅6+…=1118\frac{1}{1\cdot 4}+\frac{1}{2\cdot 5}+\frac{1}{3\cdot 6}+\ldots = \frac{11}{18}.

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Problem 5024

Prove the equality 11⋅7+13⋅9+15⋅11+…=2390\frac{1}{1\cdot 7}+\frac{1}{3\cdot 9}+\frac{1}{5\cdot 11}+\ldots = \frac{23}{90}.

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Problem 5025

Prove the equality ∑n=1∞n(2n−1)2(2n+1)2=18\sum_{n=1}^{\infty } \frac{n}{(2n-1)^{2}(2n+1)^{2}} = \frac{1}{8}.

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Problem 5026

Prove the equality ∑n=1∞2n+1n2(n+1)2=1\sum_{n=1}^{\infty } \frac{2n+1}{n^{2}(n+1)^{2}} = 1.

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Problem 5027

Prove the equality ∑n=1∞1n(n+1)(n+2)=14\sum_{n=1}^{\infty } \frac{1}{n(n+1)(n+2)} = \frac{1}{4}.

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Problem 5028

Prove the equality 11⋅3⋅5⋅7+13⋅5⋅7⋅9+15⋅7⋅9⋅11+…=190\frac{1}{1\cdot 3\cdot 5\cdot 7}+\frac{1}{3\cdot 5\cdot 7\cdot 9}+\frac{1}{5\cdot 7\cdot 9\cdot 11}+\ldots = \frac{1}{90}.

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Problem 5029

Prove the equality ∑n=1∞n2(n+1)(n+2)(n+3)(n+4)=536\sum_{n=1}^{\infty } \frac{n^{2}}{(n+1)(n+2)(n+3)(n+4)} = \frac{5}{36}.

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Problem 5030

Prove the equality ∑n=1∞1n(4n2−1)=2ln⁡2−1\sum_{n=1}^{\infty } \frac{1}{n(4n^{2}-1)} = 2\ln 2 - 1.

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Problem 5031

Prove the equality ∑n=1∞n(4n2−1)2=18\sum_{n=1}^{\infty } \frac{n}{(4n^{2}-1)^{2}} = \frac{1}{8}.

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Problem 5032

Using the identity

arctg⁡ba+n−1−arctg⁡ba+n=arctg⁡bb2+(a+n)(a+n−1), \operatorname {arctg}\frac{b}{a+n-1} - \operatorname {arctg}\frac{b}{a+n} = \operatorname {arctg}\frac{b}{b^{2}+(a+n)(a+n-1)},

prove the equality ∑n=1∞arctg⁡12n2=π4\sum_{n=1}^{\infty } \operatorname {arctg}\frac{1}{2n^{2}} = \frac{\pi }{4}.

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Problem 5033

Using the identity

arctg⁡ba+n−1−arctg⁡ba+n=arctg⁡bb2+(a+n)(a+n−1), \operatorname {arctg}\frac{b}{a+n-1} - \operatorname {arctg}\frac{b}{a+n} = \operatorname {arctg}\frac{b}{b^{2}+(a+n)(a+n-1)},

prove the equality ∑n=1∞arctg⁡1n2+n+1=π4\sum_{n=1}^{\infty } \operatorname {arctg}\frac{1}{n^{2}+n+1} = \frac{\pi }{4}.

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Problem 5034

Prove that for ∣x∣<1\left|x\right| < 1 we have the equality (1+x)(1+x2)(1+x4)(1+x8)…=11−x(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\ldots = \frac{1}{1-x}.

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Problem 5035

Prove that for ∣x∣<1\left|x\right| < 1 we have the equality x1+x+2x21+x2+4x41+x4+…=x1−x\frac{x}{1+x}+\frac{2x^{2}}{1+x^{2}}+\frac{4x^{4}}{1+x^{4}}+\ldots = \frac{x}{1-x}.

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Problem 5036

Prove that for ∣x∣<1\left|x\right| < 1 we have the equality x1−x2+x21−x4+x41−x8+…=x1−x\frac{x}{1-x^{2}}+\frac{x^{2}}{1-x^{4}}+\frac{x^{4}}{1-x^{8}}+\ldots = \frac{x}{1-x}.

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Problem 5037

Prove the equality 12tg⁡x2+14tg⁡x4+18tg⁡x8+…=1x−ctg⁡x\frac{1}{2}\operatorname {tg}\frac{x}{2}+\frac{1}{4}\operatorname {tg}\frac{x}{4}+\frac{1}{8}\operatorname {tg}\frac{x}{8}+\ldots = \frac{1}{x}-\operatorname {ctg}x.

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Problem 5038

Prove the equality ∑n=1∞122ncos⁡2x2n=1sin⁡2x−1x2\sum_{n=1}^{\infty } \frac{1}{2^{2n}\cos^{2}\frac{x}{2^{n}}} = \frac{1}{\sin^{2}x} - \frac{1}{x^{2}}.

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Problem 5039

Using the equalities:

∫01tn−1 dt=1n,∫01tn−1(1−t) dt=1n(n+1),∫01tn−1(1−t)2 dt=1n(n+1)(n+2), \int _{0}^{1} t^{n-1}\, dt = \frac{1}{n}, \quad \int _{0}^{1} t^{n-1}(1-t)\, dt = \frac{1}{n(n+1)}, \quad \int _{0}^{1} t^{n-1}(1-t)^{2}\, dt = \frac{1}{n(n+1)(n+2)},

prove that:

1−14+17−110+…=13(π3+ln⁡2). 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \ldots = \frac{1}{3}\left(\frac{\pi }{\sqrt{3}}+\ln 2\right).
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Problem 5040

Using the equalities:

∫01tn−1 dt=1n,∫01tn−1(1−t) dt=1n(n+1),∫01tn−1(1−t)2 dt=1n(n+1)(n+2), \int _{0}^{1} t^{n-1}\, dt = \frac{1}{n}, \quad \int _{0}^{1} t^{n-1}(1-t)\, dt = \frac{1}{n(n+1)}, \quad \int _{0}^{1} t^{n-1}(1-t)^{2}\, dt = \frac{1}{n(n+1)(n+2)},

prove that:

1−15+19−113+…=142(π+2ln⁡(2+1)). 1 - \frac{1}{5} + \frac{1}{9} - \frac{1}{13} + \ldots = \frac{1}{4\sqrt{2}}\left(\pi + 2\ln (\sqrt{2}+1)\right).
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Problem 5041

Prove the equality 1+12−13−14+15+16−17−18+…=π4+12ln⁡21+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\ldots = \frac{\pi }{4}+\frac{1}{2}\ln 2.

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Problem 5042

Prove the equality 1+13+15−17−19−111+113+…=5π121+\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}-\frac{1}{11}+\frac{1}{13}+\ldots = \frac{5\pi }{12}.

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Problem 5043

Prove the equality ∑n=1,3,5,…(−1)n2−18n=ln⁡(1+2)2\sum_{n=1,3,5,\ldots } \frac{(-1)^{\frac{n^{2}-1}{8}}}{n} = \frac{\ln (1+\sqrt{2})}{\sqrt{2}}.

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Problem 5044

Prove the equality ∑n=1∞1(n+α)(n+β)=1β−α∫01xα−xβ1−x dx\sum_{n=1}^{\infty } \frac{1}{(n+\alpha )(n+\beta )} = \frac{1}{\beta -\alpha }\int_{0}^{1} \frac{x^{\alpha }-x^{\beta }}{1-x}\, dx; α>−1\alpha > -1, β>−1\beta > -1, α≠β\alpha \neq \beta.

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Problem 5045

Prove the equality ∑n=0∞1(4n+1)(4n+3)=π8\sum_{n=0}^{\infty } \frac{1}{(4n+1)(4n+3)} = \frac{\pi }{8}.

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Problem 5046

Prove the equality ∑n=1∞1n(n+m)=1m(1+12+13+…+1m)\sum_{n=1}^{\infty } \frac{1}{n(n+m)} = \frac{1}{m}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{m}\right).

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Problem 5047

Prove the equality 11⋅2⋅3+13⋅4⋅5+15⋅6⋅7+…=ln⁡2−12\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\ldots = \ln 2 - \frac{1}{2}.

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Problem 5048

Prove the equality 11⋅2⋅3−13⋅4⋅5+15⋅6⋅7−…=12(1−ln⁡2)\frac{1}{1\cdot 2\cdot 3}-\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}-\ldots = \frac{1}{2}(1-\ln 2).

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Problem 5049

Prove the equality 11⋅2⋅3⋅4+14⋅5⋅6⋅7+17⋅8⋅9⋅10+…=16−14ln⁡3+π123\frac{1}{1\cdot 2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6\cdot 7}+\frac{1}{7\cdot 8\cdot 9\cdot 10}+\ldots = \frac{1}{6}-\frac{1}{4}\ln 3+\frac{\pi }{12\sqrt{3}}.

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Problem 5050

Prove the equality 112⋅32+132⋅52+152⋅72+…=π216−12\frac{1}{1^{2}\cdot 3^{2}}+\frac{1}{3^{2}\cdot 5^{2}}+\frac{1}{5^{2}\cdot 7^{2}}+\ldots = \frac{\pi^{2}}{16}-\frac{1}{2}.

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Problem 5051

Prove the equality 112⋅32⋅52+132⋅52⋅72+172⋅92⋅112+…=3π2256−19\frac{1}{1^{2}\cdot 3^{2}\cdot 5^{2}}+\frac{1}{3^{2}\cdot 5^{2}\cdot 7^{2}}+\frac{1}{7^{2}\cdot 9^{2}\cdot 11^{2}}+\ldots = \frac{3\pi^{2}}{256}-\frac{1}{9}.

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Problem 5052

Prove the equality 112⋅22⋅32+122⋅32⋅42+132⋅42⋅52+…=4π2−3916\frac{1}{1^{2}\cdot 2^{2}\cdot 3^{2}}+\frac{1}{2^{2}\cdot 3^{2}\cdot 4^{2}}+\frac{1}{3^{2}\cdot 4^{2}\cdot 5^{2}}+\ldots = \frac{4\pi^{2}-39}{16}.

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Problem 5053

Prove the equality ∑n=1∞xnn(n+1)(n+2)=34−12x−(1−x)22x2ln⁡11−x\sum_{n=1}^{\infty } \frac{x^{n}}{n(n+1)(n+2)} = \frac{3}{4}-\frac{1}{2x}-\frac{(1-x)^{2}}{2x^{2}}\ln \frac{1}{1-x}.

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Problem 5054

Prove the equality

∑n=0∞xn(2n+1)(2n+3)(2n+5)=524x−18x2+(1−x)28x2⋅R; \sum _{n=0}^{\infty } \frac{x^{n}}{(2n+1)(2n+3)(2n+5)} = \frac{5}{24x}-\frac{1}{8x^{2}}+\frac{(1-x)^{2}}{8x^{2}} \cdot R; R=12xln⁡1+x1−x, if x>0;R=1−xarctg⁡−x, if x<0. R = \frac{1}{2\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}, \text{ if } x > 0; \quad R = \frac{1}{\sqrt{-x}}\operatorname {arctg}\sqrt{-x}, \text{ if } x < 0.
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Problem 5055

Using Newton's binomial theorem, prove that

1−12+1⋅32⋅4−1⋅3⋅52⋅4⋅6+…=12. 1 - \frac{1}{2} + \frac{1\cdot 3}{2\cdot 4} - \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} + \ldots = \frac{1}{\sqrt{2}}.
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Problem 5056

Using the equality ∫01tn−1 dt=1n\int_{0}^{1} t^{n-1}\, dt = \frac{1}{n}, obtain that

2⋅4⋅6…2n3⋅5⋅7…(2n+1)=∫01(1−t2)n dt, \frac{2\cdot 4\cdot 6\ldots 2n}{3\cdot 5\cdot 7\ldots (2n+1)} = \int _{0}^{1} (1-t^{2})^{n}\, dt,

and derive the formula:

23+2⋅43⋅5x+2⋅4⋅63⋅5⋅7x2+…=∫011−t21−x+xt2 dt;∣x∣<1. \frac{2}{3}+\frac{2\cdot 4}{3\cdot 5}x+\frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}x^{2}+\ldots = \int _{0}^{1} \frac{1-t^{2}}{1-x+xt^{2}}\, dt; \quad \left|x\right| < 1.
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Problem 5057

Prove the equality 23−2⋅43⋅5+2⋅4⋅63⋅5⋅7−…=1−12ln⁡(1+2)\frac{2}{3}-\frac{2\cdot 4}{3\cdot 5}+\frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}-\ldots = 1-\frac{1}{\sqrt{2}}\ln (1+\sqrt{2}).

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Problem 5058

Prove the equality x23+2x43⋅5+2⋅43⋅5⋅7x6+…=1−1−x2arcsin⁡xx\frac{x^{2}}{3}+\frac{2x^{4}}{3\cdot 5}+\frac{2\cdot 4}{3\cdot 5\cdot 7}x^{6}+\ldots = 1-\frac{\sqrt{1-x^{2}}\arcsin x}{x}; ∣x∣<1\left|x\right| < 1.

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Problem 5059

Prove the equality 14−1⋅34⋅6+1⋅3⋅54⋅6⋅8−…=3−22\frac{1}{4}-\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}-\ldots = 3-2\sqrt{2}.

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Problem 5060

Prove the equality 12⋅4+1⋅32⋅4⋅6+1⋅3⋅52⋅4⋅6⋅8+…=12\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\ldots = \frac{1}{2}.

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Problem 5061

Noting that ∫01tn−1ln⁡t dt=−1n2\int_{0}^{1} t^{n-1}\ln t\, dt = -\frac{1}{n^{2}}; ∫01tn−1ln⁡2t dt=2n3\int_{0}^{1} t^{n-1}\ln^{2}t\, dt = \frac{2}{n^{3}}; n>0n > 0, prove the equality

1+12⋅132+1⋅32⋅4⋅152+1⋅3⋅52⋅4⋅6⋅172+…=π2ln⁡2. 1+\frac{1}{2}\cdot \frac{1}{3^{2}}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{1}{5^{2}}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\cdot \frac{1}{7^{2}}+\ldots = \frac{\pi }{2}\ln 2.
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Problem 5062

Noting that ∫01tn−1ln⁡t dt=−1n2\int_{0}^{1} t^{n-1}\ln t\, dt = -\frac{1}{n^{2}}; ∫01tn−1ln⁡2t dt=2n3\int_{0}^{1} t^{n-1}\ln^{2}t\, dt = \frac{2}{n^{3}}; n>0n > 0, prove the equality

1+12⋅133+1⋅32⋅4⋅153+…=π348+π4(ln⁡2)2. 1+\frac{1}{2}\cdot \frac{1}{3^{3}}+\frac{1\cdot 3}{2\cdot 4}\cdot \frac{1}{5^{3}}+\ldots = \frac{\pi ^{3}}{48}+\frac{\pi }{4}(\ln 2)^{2}.
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Problem 5063

Starting from the equality n!a(a+1)…(a+n)=∫01(1−t)a−1tn dt\frac{n!}{a(a+1)\ldots (a+n)} = \int_{0}^{1} (1-t)^{a-1}t^{n}\, dt, prove that

1a+xa(a+1)+x2a(a+1)(a+2)+…=∫01(1−t)a−1ext dt;a>0. \frac{1}{a}+\frac{x}{a(a+1)}+\frac{x^{2}}{a(a+1)(a+2)}+\ldots = \int _{0}^{1} (1-t)^{a-1}e^{xt}\, dt; \quad a > 0.
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Problem 5064

Prove that

1a+1a(a+1)+1a(a+1)(a+2)+…=e(1a−11!(a+1)+12!(a+2)−…). \frac{1}{a}+\frac{1}{a(a+1)}+\frac{1}{a(a+1)(a+2)}+\ldots = e\left(\frac{1}{a}-\frac{1}{1!(a+1)}+\frac{1}{2!(a+2)}-\ldots \right).
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Problem 5065

Using the identity

n!x(x+1)…(x+n−1)=n!(x−2)x(x+1)…(x+n−2)−(n+1)!(x−2)x(x+1)…(x+n−1), \frac{n!}{x(x+1)\ldots (x+n-1)} = \frac{n!}{(x-2)x(x+1)\ldots (x+n-2)} - \frac{(n+1)!}{(x-2)x(x+1)\ldots (x+n-1)},

prove the equality:

1x+2!x(x+1)+3!x(x+1)(x+2)+…=1x+2;x>2. \frac{1}{x}+\frac{2!}{x(x+1)}+\frac{3!}{x(x+1)(x+2)}+\ldots = \frac{1}{x+2}; \quad x > 2.
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Problem 5066

Prove Stirling's equality:

1x(x+1)+1+nx(x+1)(x+2)+(1+n)(2+n)x(x+1)(x+2)(x+3)+…= \frac{1}{x(x+1)}+\frac{1+n}{x(x+1)(x+2)}+\frac{(1+n)(2+n)}{x(x+1)(x+2)(x+3)}+\ldots = =1x2−nx;x>n. = \frac{1}{x^{2}-nx}; \quad x > n.
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Problem 5067

Prove that

∑n=0∞1(t+n)2=1t+12t(t+1)+2!3t(t+1)(t+2)+… \sum _{n=0}^{\infty } \frac{1}{(t+n)^{2}} = \frac{1}{t}+\frac{1}{2t(t+1)}+\frac{2!}{3t(t+1)(t+2)}+\ldots
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Problem 5068

Prove that for γ>α>0\gamma > \alpha > 0, ∣x∣<1\left|x\right| < 1 we have the equality

F(α,β,γ;x)=Γ(γ)Γ(α)Γ(γ−α)∫01uα−1(1−u)γ−α−1(1−xu)−β du, F(\alpha , \beta , \gamma ; x) = \frac{\Gamma (\gamma )}{\Gamma (\alpha )\Gamma (\gamma -\alpha )} \int _{0}^{1} u^{\alpha -1}(1-u)^{\gamma -\alpha -1}(1-xu)^{-\beta }\, du,

where

F(α,β,γ;x)=1+α⋅β1⋅γx+α(α+1)1⋅2β(β+1)γ(γ+1)x2+… F(\alpha , \beta , \gamma ; x) = 1+\frac{\alpha \cdot \beta }{1\cdot \gamma }x+\frac{\alpha (\alpha +1)}{1\cdot 2}\frac{\beta (\beta +1)}{\gamma (\gamma +1)}x^{2}+\ldots

(Gauss's hypergeometric series).

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Problem 5069

Prove that for Re⁡(γ−α−β)>0\operatorname {Re}(\gamma -\alpha -\beta ) > 0 we have the equality

F(α,β,γ;1)=Γ(γ)Γ(γ−α−β)Γ(γ−α)Γ(γ−β) F(\alpha , \beta , \gamma ; 1) = \frac{\Gamma (\gamma )\Gamma (\gamma -\alpha -\beta )}{\Gamma (\gamma -\alpha )\Gamma (\gamma -\beta )}

(Re⁡\operatorname {Re} denotes the real part of a complex number).

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Problem 5070

Prove the equality 1+abx+a(a+1)b(b+1)x2+…=1+ax∫01(1−u)b−1(1−ux)a−1 du1+\frac{a}{b}x+\frac{a(a+1)}{b(b+1)}x^{2}+\ldots = 1+ax\int_{0}^{1} \frac{(1-u)^{b-1}}{(1-ux)^{a-1}}\, du; b>0b > 0.

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Problem 5071

Prove the equality 12+(12)2⋅14+(1⋅32⋅4)2⋅16+…=2π\frac{1}{2}+\left(\frac{1}{2}\right)^{2}\cdot \frac{1}{4}+\left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\cdot \frac{1}{6}+\ldots = \frac{2}{\pi }.

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Problem 5072

Prove the equality (12)2+3⋅(12⋅4)2+5⋅(1⋅32⋅4⋅6)2+…=1−2π\left(\frac{1}{2}\right)^{2}+3\cdot \left(\frac{1}{2\cdot 4}\right)^{2}+5\cdot \left(\frac{1\cdot 3}{2\cdot 4\cdot 6}\right)^{2}+\ldots = 1-\frac{2}{\pi }.

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Problem 5073

Prove the equality φ(x)+φ(x+π2)=4φ(2x)\varphi (x)+\varphi \left(x+\frac{\pi }{2}\right) = 4\varphi (2x), where

φ(x)=∑n=−∞∞1(x−nπ)2−1sin⁡2x. \varphi (x) = \sum _{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}} - \frac{1}{\sin ^{2}x}.
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Problem 5074

Using the result of problem 5073 (the equality φ(x)+φ(x+π2)=4φ(2x)\varphi (x)+\varphi \left(x+\frac{\pi }{2}\right) = 4\varphi (2x), where φ(x)=∑n=−∞∞1(x−nπ)2−1sin⁡2x\varphi (x) = \sum_{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}} - \frac{1}{\sin^{2}x}), show that

1sin⁡2x=∑n=−∞∞1(x−nπ)2. \frac{1}{\sin ^{2}x} = \sum _{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}}.
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Problem 5075

By integration, obtain Euler's equality

ctg⁡x=1x+∑n≠0(1x−nπ+1x+nπ)=1x+∑n=1∞2xx2−n2π2. \operatorname {ctg}x = \frac{1}{x}+\sum _{n \neq 0}\left(\frac{1}{x-n\pi }+\frac{1}{x+n\pi }\right) = \frac{1}{x}+\sum _{n=1}^{\infty } \frac{2x}{x^{2}-n^{2}\pi ^{2}}.
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Problem 5076

By integrating and passing from logarithms to numbers, prove the equality:

sin⁡x=x∏n=1∞(1−x2n2π2).(Euler.) \sin x = x\prod _{n=1}^{\infty }\left(1-\frac{x^{2}}{n^{2}\pi ^{2}}\right). \qquad \text{(Euler.)}
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Problem 5077

Prove Euler's equality giving the expansion into an infinite series of fractions of the function tg⁡x=∑n=0∞2x(2n+12π)2−x2\operatorname {tg}x = \sum_{n=0}^{\infty } \frac{2x}{\left(\frac{2n+1}{2}\pi \right)^{2}-x^{2}}.

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Problem 5078

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 1sin⁡x=1x+∑n=1∞(−1)n⋅2xx2−n2π2\frac{1}{\sin x} = \frac{1}{x}+\sum_{n=1}^{\infty } \frac{(-1)^{n}\cdot 2x}{x^{2}-n^{2}\pi^{2}}.

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Problem 5079

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 1cos⁡x=∑n=1∞(−1)n(2n−1)πx2−(n−12)2π2\frac{1}{\cos x} = \sum_{n=1}^{\infty } \frac{(-1)^{n}(2n-1)\pi }{x^{2}-\left(n-\frac{1}{2}\right)^{2}\pi^{2}}.

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Problem 5080

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 11−e−x−1x−12=∑n=1∞2xx2+4π2n2\frac{1}{1-e^{-x}} - \frac{1}{x} - \frac{1}{2} = \sum_{n=1}^{\infty } \frac{2x}{x^{2}+4\pi^{2}n^{2}}.

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